Question

The final volume of buffer solution must be 100.00 mL and the final concentration of the weak acid must be 0.100 M.

Based on this information, what mass of solid conjugate base should the student weigh out to make the buffer solution with a pH=7.0 (in grams)? Based on this information, what volume of acid should the student measure to make the 0.100 M buffer solution (in mL)?

My weak acid is Sodium dihydrogen phosphate monohydrate, Ka= 6.23X10^-8, 2.00M and my conjugate base is Disodium hydrogen phosphate heptahydrate, Na2HPO₄*7H₂O

Answer 1

Sol:-

Given final concentration of weak acid i.e [NaH₂PO₄.H₂O] = 0.100 M

Also final volume = 100.00 mL = 0.100 L

K_a of NaH₂PO₄.H₂O = 6.23 x 10^-8

therefore

pK_a = - log K_a = - log 6.23 x 10^-8 = - ( - 7.20) = 7.20

so

pK_a of NaH₂PO₄.H₂O = 7.20 .

given that pH of acidic buffer solution = 7.0

Now by using Henderson-Hasselbalch equation , we have

pH = pK_a + log [conjugate base ] / [ Acid ]    

pH = pK_a + log [ Na₂HPO₄.7H₂O] / [ NaH₂PO₄.H₂O]

7.0 = 7.20 + log [ Na₂HPO₄.7H₂O] / [ NaH₂PO₄.H₂O]

log [ Na₂HPO₄.7H₂O] / [ NaH₂PO₄.H₂O] = 7.0 - 7.20

log [ Na₂HPO₄.7H₂O] / [ NaH₂PO₄.H₂O] = -0.20

[ Na₂HPO₄.7H₂O] / [ NaH₂PO₄.H₂O] = 10^-0.20

[ Na₂HPO₄.7H₂O] / [ NaH₂PO₄.H₂O] = 0.63

[ Na₂HPO₄.7H₂O] = 0.63 x [ NaH₂PO₄.H₂O]

[ Na₂HPO₄.7H₂O] = 0.63 x 0.100 M

[ Na₂HPO₄.7H₂O] = 0.063 M

Number of moles of Na₂HPO₄.7H₂O / final volume in litre = 0.063 M
( because Molarity = number of moles of solute / volume of solution in litre )

therefore

Number of moles of Na₂HPO₄.7H₂O = 0.063 M x 0.100 L

Number of moles of Na₂HPO₄.7H₂O = 0.0063 mol

Also Number of moles of substance = mass of substance in gram / gram molar mass of substance .

and gram molar mass of Na₂HPO₄.7H₂O = 268 g/mol

therefore

mass of Na₂HPO₄.7H₂O = moles of Na₂HPO₄.7H₂O x gram molar mass of Na₂HPO₄.7H₂O

mass of Na₂HPO₄.7H₂O = 0.0063 mol x 268 g/mol

mass of Na₂HPO₄.7H₂O = 1.69 g

Hence mass of Na₂HPO₄.7H₂O used = 1.69 g

Number of moles of weak acid i.e NaH₂PO₄.H₂O = Molarity of NaH₂PO₄.H₂O x final volume in litre

Number of moles of weak acid i.e NaH₂PO₄.H₂O = 0.100 M x 0.100 L = 0.01 mol

given initial concentration of NaH₂PO₄.H₂O = 2.00 M

Molarity = number of moles of solute / volume of solution in litre

therefore

Volume of solution = number of moles of NaH₂PO₄.H₂O / Molarity of NaH₂PO₄.H₂O

Volume = 0.01 mol / 2.00 mol/L

Volume = 0.005 L

Volume = 5.00 mL

Hence Volume of NaH₂PO₄.H₂O used = 5.00 mL